Question

A neutron collides elastically with a helium nucleus (at rest initially) whose mass is four times that of the neutron. The helium nucleus is observed to rebound at an angle '2 = 43° from the neutron's initial direction. The neutron's initial speed is 5.4 105 m/s. Determine the angle at which the neutron rebounds, '1, measured from its initial direction.
°
What is the speed of the neutron after the collision?
m/s
What is the speed of the helium nucleus after the collision?

Answer 1

Conservation of momentum means that the initial momentum of the neutron will be split between it and the helium atom after the collision. Call that initial direction the x direction, then after collision all that momentum goes into the x-direction components of momentum for the neutron and the helium atom. The initial y-momentum was zero, so the sum of the post-collision components of y momentum will be zero.

Elastic means that the kinetic energy is conserved, so that the (1/2)mv^2 of the neutron initially will be shared by the neutron and the helium atom afterward.

Answer 2

How would I be able to find theta though? And since it is elastic the neutron and helium would have the same speed?

Answer 3

Elastic only means energy is conserved, not that they have the same speed.

You get theta by looking at the components of velocity in the x and the y direction. Rebounding at a velocity v at a theta=43o angle from the x direction would mean helium having an x component of velocity v cos (43) and a y component of v sin(43). Try to work it out and I will spend a little time on it, too.

Make life easier by setting m1 of neutron = 1, m2 of helium=4. Their ratios should be what is important in this problem, not their absolute masses..

Answer 4

Well, it gets messy. You have three equations: two for momentum (x, y directions) and one for energy. You know the masses, and the initial velocity, and one final angle (43o), but you have three unknowns: theta and the resulting two velocities. I have not been able to boil them down yet.

x momentum: m v1 = M v2 cos(43) + m v1' cos(theta)
y momentum: 0 = M v2 sin(43) +m v1' sin(theta)
energy: (1/2)m v1^2 = (1/2) M v2^2 + (1/2) m v1'^2

Using M/m=4 and rearranging, the energy equation
gives v1^2 - v1'^2 = 4 v2^2, the energy lost by the neutron goes to He.

The y momentum equation shows that
sin(theta) = - M v2 sin(43) / m v1' = (-4 v2 / v1') sin(43)
so that theta is directed somewhat below the horizontal.

Brute force with the momentum equations will let you get rid of the terms with M v2 in them and get v1' and theta, probably, but I am stopping here!

Answer 5

Thanks for all the help!